Re: Absolute value equation | x - 4 | = x/3 + 2

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 01/29/05 

Date: Sat, 29 Jan 2005 14:32:00 -0600

On Sat, 29 Jan 2005 17:38:01 +0000 (UTC), George Cox
<george_coxanti@spambtinternet.com.invalid> wrote:

>"David C. Ullrich" wrote:
>>
>> On Sat, 29 Jan 2005 05:16:07 +0000 (UTC), George Cox
>> <george_coxanti@spambtinternet.com.invalid> wrote:
>>
>> >Al wrote:
>> >>
>> >> Can anyone solve this?
>> >>
>> >> | x - 4 | = x/3 + 2
>> >
>> > x - 4 = x/3 + 2 or 4 - x = x/3 + 2
>>
>> This gave the right answer, but only by luck,
>> there's an important detail missing from the
>> way you set it up. You meant to say
>>
>> [x - 4 >= 0 and x - 4 = x/3 + 2]
>>
>> or
>>
>> [x - 4 <= 0 and 4 - x = x/3 + 2]
>
>You're right, I was lucky. On the day I answered the question
>
> [x - 4 >= 0] or [x - 4 <= 0]
>
>was true. Had it not been I would have misled the OP. Giggle.

Um, the fact that [A or not A] is always true does not
make ({A and P] or [(not A) and Q]) equivalent to (P or Q).

Instead of worrying about the logic, let's look
at a similar question where your reasoning
gives an incorrect answer. Let's solve

  |x| = x/2 - 1.

We "rewrite" that as "x = x/2 - 1 or -x = x/2 - 1",
giving two solutions x = -2 and x = 2/3. Neither
of which is actually a solution to |x| = x/2 - 1.

You got lucky in that the solution to x - 4 = x/3 + 2
satisfied x - 4 >= 0, and similarly for the other
solution.

Giggle, indeed.

************************

David C. Ullrich

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